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Let $\vec{a}$ and $\vec{b}$ be vectors from $V_3$. Suppose, that $|\vec{a}| = 1$, $|\vec{b}|=2$ and the angle between $\vec{a},\vec{b}$ is $\frac{\pi}{3}$. Use the properties of scalar product and compute $|\vec{a}-2\vec{b}|$.

I'm totaly lost in that.

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The properties you're expected to use are bilinearity of the scalar product (meaning that you can "multiply out" expressions with scalar products by pretending that it behaves just like the ordinary product of real numbers) and the fact that $|\vec c|=\sqrt{\vec c\cdot\vec c}$. Now take $\vec c=\vec a-2\vec b$ and work away... Does that help? –  user108903 Jan 20 '13 at 10:33

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Hint: $|\vec{a}-2\vec{b}|$ is the square root of $(\vec{a}-2\vec{b}) \cdot (\vec{a}-2\vec{b})$.

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$\lvert \vec{a} - 2 \vec{b} \rvert^2 = \langle \vec{a} – 2\vec{b},\vec{a} – 2\vec{b} \rangle = \langle \vec{a}, \vec{a} \rangle - 4 \langle \vec{a}, \vec{b} \rangle + 4 \langle \vec{b},\vec{b} \rangle$.

Now, can you compute $\langle \vec{a},\vec{b} \rangle$, knowing $\angle (\vec{a},\vec{b}) = \tfrac{\pi}{3}$?

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No, I'm afraid I can not. –  user50222 Jan 20 '13 at 11:12
    
@user50222: There is this identity. You have to know $\cos (\tfrac{\pi}{3})$. –  k.stm Jan 20 '13 at 11:20

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